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A clockwise engine Carnot cycle has four reversible steps. At hot temperature $T_h$, isothermal expansion absorbs $Q_h=T_h\Delta S>0$ and does positive work by the gas. Adiabatic expansion then cools it to $T_c$, with zero heat and positive work by the gas. Isothermal compression at $T_c$ emits $Q_c=-T_c\Delta S$ and requires work on the gas. Finally adiabatic compression returns it to $T_h$, with zero heat and work on the gas.

For an ideal gas, the $p,V$ isotherms obey $pV=\text{constant}$ and the adiabats $p{V}^{\gamma }=\text{constant}$, the latter being steeper. In the $T,S$ plane, the isotherms are horizontal and the adiabats vertical, forming a clockwise rectangle. Heat occurs only on the horizontal legs, while work occurs on all four $p,V$ legs.Reversing the arrows gives a refrigerator, reversing the heat and work signs.

Let $e=(1,0)$ and defineThen $\Vert g(x)-x\Vert \le 1$ everywhere, but $0$ is not in the image of a function. The map is discontinuous at $0$, giving a bounded-displacement map which is not surjective.

For arbitrary $y\in {\mathbb{R}}^2$, the continuous map $g_y(x)=g(x)-y$ satisfies $\Vert g_y(x)-x\Vert \le K+\Vert y\Vert$. Applying the preceding Brouwer fixed-point theorem argument with this positive bound gives $g_y(t)=0$. Hence $g(t)=y$, and $g$ is surjective.

Let $D$ be the closed unit disc and setThe displacement bound gives $\Vert f(x)\Vert \le 1$, so $f:D\to D$ is continuous. The Brouwer fixed-point theorem applies because $D$ is nonempty, compact and convex. At a fixed point $x_{\ast }$, $x_{\ast }=x_{\ast }-g(K{x}_{\ast })/K$. Therefore $\boxed{t=K{x}_{\ast },g(t)=0}$.

Put $K^{\prime }=K(1+r{)}^{t_0-T}$. At the choice time,Buy a European call option of strike $K$ and maturity $T$, together with a European put option of strike $K^{\prime }$ and maturity $t_0$. If the call is selected, the short-maturity put expires worthless. If the put is selected, sell the call and combine its value with the short-maturity put's cash payment; put-call parity makes this exactly enough to buy the maturity-$T$ put. This self-financing replication provesIt requires tradability of the priced options at the choice time, but not completeness of every other market claim. The identity also covers $t_0=0$ and $t_0=T$.

The difference of the terminal European call option and European put option payoffs is $S_T^1-K$. Replicating that difference by one share and a borrowing position gives conditional put-call parityThus the call is more valuable exactly when the right side is positive. With a put selected at equality, the chooser option payoff isAt equality both choices have the same price, even though their eventual payoffs can differ. Put-call parity assumes no dividends or other intermediate stock cash flows, as in the stated asset-price model.

For a finite-state, finite-horizon discrete-time frictionless market with a strictly positive numéraire, the fundamental theorem of asset pricing states no arbitrage if and only if there exists an equivalent martingale measure under which all discounted asset prices are martingales. Equivalence means the same null events as the physical measure. In this setting market completeness is equivalent to uniqueness of the equivalent martingale measure, and a replicable payoff $H$ has price $S_t^0{\mathbb{E}}_Q[H/S_T^0\mid {\mathcal{F}}_t]$. Without finite-state hypotheses, stronger integrability and closure conditions are needed; this is the standard multi-period course version.

The almost-sure bound gives ${\tilde{\theta }}_n-{\hat{\theta }}_n=o_{\mathbb{P}}(n^{-1/2})$, so Slutsky theorem transfers both consistency and the asymptotic normal law to ${\tilde{\theta }}_n$. The continuous mapping theorem and delta method then giveIf ${\psi }^{\prime }({\theta }_0)=0$, the latter is the Dirac measure at zero, conventionally a degenerate normal distribution; a nondegenerate second-order limit would need a different normalization and extra smoothness.

A maximum-likelihood estimate ${\hat{\theta }}_n$ is a measurable maximizer of ${\ell }_n(\theta )={\sum }_i\log f(X_i,\theta )$ over $\Theta$. Under identifiability, consistency conditions, an interior true parameter, sufficient differentiability and finite positive Fisher information, the standard limits areThe variance in the normal distribution is the reciprocal of the per-observation Fisher information, not of $nI$. Strong consistency needs additional hypotheses if it is to replace convergence in probability; the regular asymptotic result stated here does not require claiming it.

The per-observation Fisher information is $I(\theta )={\mathbb{E}}_{\theta }[({\partial }_{\theta }\log f(X_1,\theta ){)}^2]$. Sinceand the integral of $f_{\theta \theta }$ is the second derivative of $1$, the regularity assumptions giveFor independent observations the total Fisher information is $nI(\theta )$, because centered individual score functions have zero cross terms.

The score function isWith common parameter-independent support and justified differentiation under the integral, ${\mathbb{E}}_{\theta }[{\partial }_{\theta }\log f(X_1,\theta )]=\int {\partial }_{\theta }f(x,\theta )dx={\partial }_{\theta }1=0$. Linearity then gives $\boxed{{\mathbb{E}}_{{\theta }_0}{S}_n({\theta }_0)=0}$. These regularity conditions exclude, for example, moving-support uniform models.

A renewal process has independent identically distributed, nonnegative, almost surely finite inter-arrival times $X_i$, epochs $S_0=0,S_n={\sum }_{i=1}^n{X}_i$, and count $N(t)=\max \{n:S_n\le t\}$. Allowing zero inter-arrivals is harmless provided $\mathbb{P}(X_1>0)>0$. The hypothesis $\mathbb{E}{X}_1>0$ ensures there is $\epsilon >0$ with $p=\mathbb{P}(X_1\ge \epsilon )>0$.

The strong law of large numbers for the indicators gives $S_n\ge \epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\to \infty$, so $N(t)$ is finite for finite $t$. Conversely every fixed $S_n$ is finite, so eventually $N(t)\ge n$. On their countable common probability-one event, this proves $\boxed{N(t)\to \infty \text{ as }t\to \infty \text{ almost surely}}$.

For fixed $t$, put $L=\lfloor t/\epsilon \rfloor$. The hinted comparison process has epochs $S_n^{\prime }=\epsilon {\sum }_{i=1}^n{1}_{X_i\ge \epsilon }\le S_n$, hence $N(t)\le N^{\prime }(t)$. Equivalently,If $p<1$, for $n>L$ this is bounded by $C_{t,p}{n}^L(1-p{)}^n$, by summing the first $L+1$ binomial distribution. Thus ${\sum }_n{e}^{\theta n}\mathbb{P}(N(t)=n)<\infty$ for $0<\theta <-\log (1-p)$. If $p=1$, $N(t)\le L$ deterministically. Therefore a strictly positive exponential moment exists for every fixed finite $t$, with the same allowed $\theta$ for all such $t$. Almost-sure finiteness of the inter-arrivals is implicit in a proper renewal process; allowing an infinite inter-arrival with positive probability would invalidate the divergence claim.

In a neutral Moran model, a population has $N$ labelled individuals; reproduction copies one individual's type to a uniformly chosen other individual, keeping population size fixed. One convenient time convention puts independent rate-$1/2$ Poisson arrows on each ordered pair of distinct individuals. Backwards in time, two sampled ancestral lineages merge whenever one copies the other, at total rate $1$ per unordered pair. In the convention where each individual reproduces at rate $1$, this pair rate is $2/(N-1)$ if the parent cannot replace itself, and time must be rescaled accordingly.

Kingman's coalescent on $n$ labels is the continuous-time Markov chain on partitions of $\{1,\dots ,n\}$, initially all singleton blocks, in which each pair of distinct blocks merges at rate $1$. With $j$ blocks, the waiting time is exponential of rate $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$, and the merging pair is uniform. Kingman's infinite coalescent is the consistent partition process on $\mathbb{N}$ whose restriction to each finite label set has this law; consistency and the extension theorem construct it. These time conventions are part of the definition.

Let $B_t^{(n)}$ be the number of blocks in the first $n$ labels. The expected time to reach at most $m$ blocks isMarkov inequality gives $\mathbb{P}(B_t^{(n)}>m)\le 2/(mt)$. As $n$ increases these counts increase to the total number of infinite-coalescent blocks, so the same bound holds for that number. Letting $m\to \infty$ shows it is finite almost surely for each $t>0$. Apply this simultaneously to positive rational times; monotonicity of the block count then covers every positive real time. Hence Kingman's infinite coalescent comes down from infinity almost surely.

The assertion is false. On $([0,1],\mathcal{B},\text{Lebesgue measure})$, put $X_n=n{1}_{(0,1/n)}$. Then $\mathbb{E}|X_n|=1$ for every $n$, so the family is $L^1$ bounded. For every $M$, choose $n>M$; then $\mathbb{E}[|X_n|1_{|X_n|>M}]=1$. The tail criterion for uniform integrability fails.

$L^p$ boundedness means ${\sup }_n\Vert X_n{\Vert }_p<\infty$, with the essential-supremum norm for $p=\infty$. For finite $p>1$, let $C={\sup }_n\mathbb{E}|X_n{|}^p<\infty$. On $|X_n|>M$, $|X_n|\le M^{1-p}|X_n{|}^p$, soThis proves uniform integrability. For an $L^{\infty }$ bounded family the tails are identically zero beyond the common essential bound.

For conjugate exponents $1<p,q<\infty$, $1/p+1/q=1$, Hölder's inequality is $\boxed{\mathbb{E}|XY|\le \Vert X{\Vert }_p\Vert Y{\Vert }_q}$. If either finite norm is zero, the corresponding variable is zero almost surely. Otherwise normalize $u=|X|/\Vert X{\Vert }_p$, $v=|Y|/\Vert Y{\Vert }_q$. Young inequality $uv\le u^p/p+v^q/q$ follows by minimizing the right side minus $uv$ in $u\ge 0$, whose derivative is $u^{p-1}-v$. Integration gives $\mathbb{E}(uv)\le 1/p+1/q=1$, proving the result. Infinite norms give the usual extended-value inequality whenever the right side is unambiguous. At the endpoints, $\mathbb{E}|XY|\le \Vert X{\Vert }_1\Vert Y{\Vert }_{\infty }$ follows from the almost-sure bound for $Y$, with the symmetric version for $p=\infty$.

A family $(X_n)$ of integrable random variables is uniformly integrable whenEquivalently the integrals of their tails vanish uniformly. This includes uniform $L^1$ boundedness, since $\mathbb{E}|X_n|\le M+\mathbb{E}[|X_n|1_{|X_n|>M}]$.

Let $F(y)=y_1^2{y}_2+y_2^2{y}_3+y_3^2{y}_1$. The proper transform isIndeed, in the blowup chart $y_i\ne 0$, write $x=ty$ with $y_i=1$. Homogeneity gives $F(x)=t^{3}F(y)$. Away from the exceptional divisor $t=0$, the equation is $F(y)=0$, and its closure retains that equation, removing the exceptional factor $t^3$. Every point with $t=0,F(y)=0$ is approached algebraically along the line $(ty,[y])$, so no component in this displayed set is extraneous.

For $x\ne 0$, the fibre is the single point $[y]=[x]$, and the projection is an isomorphism over $Y\setminus \{0\}$. Over $0$ the fibre is the projective plane cubic $F(y)=0$. In characteristic zero this cubic is smooth: a coordinate zero in the three derivative equations would force all coordinates zero; with all nonzero, multiplying $2y_1{y}_2=-y_3^2$, $y_1^2=-2y_2{y}_3$, $y_2^2=-2y_3{y}_1$ gives $1=-8$, impossible. Thus the exceptional fibre is a smooth genus-one plane cubic. The fibre description itself does not require this optional smoothness conclusion; characteristic $3$ requires separate treatment.